# Calculating volume with a triple integral

I have the following problem. I need to find the volume of the shape that is bounded up by $$u(x,y)=13+x^2$$ and down by $$v(x,y)=7x+4y$$ and from its sides by the cylinder $$x^2+y^2=1.$$ Now I know I need to do to the integral $\displaystyle \iint\limits_D \left(u(x,y)-v(x,y) \right)dxdy$

But I dont really know how to define $D$, I know that is by the projection in $XY$ plane.

I tried to draw it in the $XY$ plane, drew the cylinder and the plane $z=7x+4y$. But I don't really know where to go next.

Some tips will really help me! :)

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anyway. the projection is the cylinder, and after moving to polar cordinates the integral is easy. I just had a calculation error that made me think it was more complicated... answer is $13\pi + \pi/4$ if anyone tried.
Neither $u(x,y)$ nor $v(x,y)$ define a curve, which would usually be the boundary of a region. You should have something like $u(x,y)=k$ for some given $k$. Then there is no sign of $z$ in the problem except when you talk about volume and a cylinder. As there is no limitation on $z$, the volume is infinite. Maybe what you really want is $\displaystyle \iint\limits_D \left(u(x,y)-v(x,y) \right)dxdy$ with $D$ being the region inside the circle $x^2+y^2=1?$ That is at least a problem that can be solved. You would set it up as $$\int_{-1}^1 \int _{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}13+x^2-7x-4y\; dy\; dx$$